Question: The sides of triangle $PQR$ are tangent to a circle with center $C$ as shown. Given that $\angle PQR = 65^\circ$ and $\angle QRC = 30^\circ$, find $\angle QPR$, in degrees.

[asy]
unitsize(1.0 cm);

pair Q, P, R, C;

Q = (2.43,3.46);
P = (0,0);
R = (4.43,0);
C = incenter(Q,P,R);

draw(Q--P--R--cycle);
draw(incircle(Q,P,R));
draw(R--C);

label("$Q$", Q, N);
label("$P$", P, SW);
label("$R$", R, SE);
label("$C$", C, N);

[/asy]
Solution: The circle with center $C$ is the incircle of $\triangle PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.

From $\angle QRC = 30^\circ$, we find that $\angle QRP = 60^\circ$ because $RC$ is an angle bisector.

The sum of the measures of the internal angles of a triangle is $180^\circ$, so

\begin{align*}
\angle QPR &= 180^\circ - \angle PQR - \angle QRP \\
&= 180^\circ - 65^\circ - 60^\circ.
\end{align*}This yields $\angle QPR = \boxed{55^\circ}$.